# Algebraic manipulation with indices

The question is:

For a>0 and $$\sqrt{a}+\frac{1}{\sqrt{a}}=3$$, find the value of $$a\sqrt{a}+\frac{1}{a\sqrt{a}}$$

So I first squared the given equation and got:

$$a+\frac{1}{a}+2=9$$ $$a+\frac{1}{a}=7$$

Then to get the form of $$a\sqrt{a}+\frac{1}{a\sqrt{a}}$$:

$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$$

And I just got stuck right here because I didn't really know what to do with the $$\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$$. So I looked into the solutions and apparently it's

$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\sqrt{a}+\frac{1}{\sqrt{a}}$$

I'm not sure how those two are equal...

Note that $$\frac{\sqrt a}a = \frac{\sqrt a}{\sqrt a\cdot\sqrt a} = \frac1{\sqrt a}\\ \frac a{\sqrt a} = \frac{\sqrt a\cdot \sqrt a}{\sqrt a} = \sqrt a$$ From this we get that "those two" are indeed equal.
The top equality might be worth remembering specifically, because some people prefer to write $$\frac1{\sqrt 2}$$, while some people prefer to write $$\frac{\sqrt2}2$$. Knowing that it's the same number, without having to check every time, will make reading them faster.
Let $$x=\sqrt{a}$$ then $$x+\frac{1}{x}=3$$ and $$\\a\sqrt{a}+\frac{1}{a\sqrt{a}}=x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3x\cdot\frac{1}{x}(x+\frac{1}{x})=27-3\cdot3=18$$
$$\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}=\sqrt{a}+\frac{1}{\sqrt{a}}=3$$ !
• $3!$ is not the right answer (be careful with exclamation marks). – Arthur Nov 7 '18 at 12:41